If α and β be the roots of the equation x² - 2x + 2 = 0, then the least value of n for which (α/β)ⁿ

Q: Why do we get complex roots? I thought roots were always real numbers.

That's a common misconception! Real numbers are just a subset of the complex plane. When the discriminant $b^2 - 4ac$ is negative, the quadratic formula naturally leads us into the complex domain. Embrace it—it's where the most beautiful symmetries in mathematics live!

Q: Is there a faster way than rationalizing the denominator?

You could convert to polar form using $r(\cos\theta + i\sin\theta)$, but for simple values like $1+i$ and $1-i$, rationalization is often faster and less prone to calculation errors. Always choose the path that minimizes algebraic friction.

Q: Why is the answer $n=4$? Can't $n$ be $0$?

Great catch! While $z^0 = 1$ is mathematically true for any non-zero $z$, in the context of JEE problems, we are almost always looking for the least *positive* integer $n$. Always check the problem constraints!

Question

If $\alpha$ and $\beta$ be the roots of the equation $x^2 - 2x + 2 = 0$, then the least value of $n$ for which $(\frac{\alpha}{\beta})^n = 1$ is :

Options:

(A) $2$

(B) $3$

(C) $4$

(D) $5$

Sigma Percentile · Accepted Answer

### Analyzing the Quadratic Equation

The quadratic equation $x^2 - 2x + 2 = 0$ serves as a gateway into the complex plane. To solve for $x$, we utilize the quadratic formula:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

Here, the coefficients are $a = 1$, $b = -2$, and $c = 2$.

### The Discriminant and Complex Roots

We calculate the discriminant $D$ to determine the nature of the roots:

$$D = b^2 - 4ac = (-2)^2 - 4(1)(2) = 4 - 8 = -4$$

Since $D < 0$, we enter the domain of complex numbers. Using $i = \sqrt{-1}$, we find the square root of the discriminant:

$$\sqrt{-4} = 2i$$

Applying this to the quadratic formula, we obtain the roots:

$$x = \frac{2 \pm 2i}{2} = 1 \pm i$$

Thus, our roots are $\alpha = 1 + i$ and $\beta = 1 - i$.

### Symmetry in the Argand Plane

When plotted on the Argand plane, $\alpha = 1 + i$ and $\beta = 1 - i$ exhibit perfect symmetry. They are **complex conjugates**, mirroring each other across the real axis. This property is a fundamental characteristic of polynomials with real coefficients.

### Simplifying the Ratio

We now evaluate the ratio $\frac{\alpha}{\beta} = \frac{1 + i}{1 - i}$. To simplify, we multiply the numerator and denominator by the conjugate of the denominator, $1 + i$:

$$\frac{1 + i}{1 - i} 	imes \frac{1 + i}{1 + i} = \frac{(1 + i)^2}{1^2 - i^2}$$

Expanding the numerator and denominator:

$$\frac{1 + 2i + i^2}{1 - (-1)} = \frac{1 + 2i - 1}{2} = \frac{2i}{2} = i$$

The expression simplifies elegantly to the imaginary unit $i$.

### The Final Condition

The problem requires finding the least positive integer $n$ such that:

$$\left( \frac{\alpha}{\beta} ight)^n = 1 \Rightarrow i^n = 1$$

Recalling the cyclic nature of the powers of $i$:

$$i^1 = i, \quad i^2 = -1, \quad i^3 = -i, \quad i^4 = 1$$

The cycle repeats every four powers. Therefore, the smallest positive integer $n$ that satisfies the condition is **$n = 4$**.

Animated Solution for Mathematics - Complex Numbers: If $α$ and $β$ be the roots of the equation $x^{2} - 2 x + 2 = 0$ , then the least value of $n$ for which $(\frac{α}{β})^{n} = 1$ is :

Select Answer:

Visualized Solution

The Quadratic Equation

The Quadratic Formula

Substituting Coefficients

Simplifying the Discriminant

Finding the Complex Roots

Roots on the Complex Plane

Setting up the Ratio

Substituting into the Ratio

Rationalizing the Denominator

Expanding the Terms

The Simplified Ratio

Geometric Meaning of the Ratio

Applying the Power Condition

Powers of $i$

The Least Value of $n$

The Sigma Insight: Algebraic Operations on Complex Numbers

Analyzing the Quadratic Equation

The Discriminant and Complex Roots

Symmetry in the Argand Plane

Simplifying the Ratio

The Final Condition

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Animated Solution for Mathematics - Complex Numbers: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If α and β be the roots of the equation x2−2x+2=0, then the least value of n for which (βα​)n=1 is :